How do i even begin to understand what it's asking me?

how do i even begin to understand what it's asking me?

gotta find the asymptotes nigga

how?

are you actually retarded

Fucken hell, it is not like, the task nearly gives you the solution. Inequality of y

If you've can't solve that, you shouldn't be here OP

I don't get it.
What do they mean by "solve the inequality?"

to solve the inequality of 1/x

but what the fuck does that even mean?
how did you get those answers?
pls help

Except that you gave the range of all the bad solutions.

what the fuck man

dude 1/x= 2
1=x*2
x=1/2

HOW IS THAT HARD GO TO SCOOL AND LEAVE B UNDERAGE

my bad

1/x is your function, represented by the line on the graph, you want to know the range of x which is below the line y=2, so you basically draw the line y=2, and you put together all teh solutions :
it should be ]-infinite;0[ U ]1/2;+infinite[

I put away 0, because your function 1/x is defined on R*, 0 can't be a solution.

1/x

x is an element if R except for all Numbers of R between 0 and 0.5 but 0.5 is part of the solution.

It's asking what values of x cause 1/x to be less than or equal to 2.

"Solve the inequality 1/x

What about the left side lol

Except that you can't multiply by x, because you don't know if it is positive, so it fucks up the inequation. With your method you completely forgot the solutions ]-infinite;0[

omg THANK YOU so much kind user
I really mean it

but it's also true for x

Just need to put the 0 away because 1/x is defined on R*, but otherwise finally an user with a brain

except that you can just say x=-x and look at the solution.......

Can you math fags here help me with my calculus shit?
Whats the solution for ∫∫sinydydx with (0,4) and (sqrtx,2) integration limits?

How about you grow a fucking spine and tell your instructor that no one outside of engineers and instructors uses this shit in the real world, take your D+ and get a career in the real world actually producing something and earning a fucking salary?

I think it wants you to draw the graph of y=2 and then find all the places where the graph of 1/x is below that line.

a double integral or what?

It's easy. It's not fucking rocket appliances.

Yeah. Dubz

how about you stop being a retarded faggot.

>inequalty
> "="

> how about you substitute = to

draw the horizontal line y=2. All of the red line, y=1/x above y=2 is wrong. Everything at the blue line and below is right.

when you plug in any number for x in 1/x if its less than or equal to 2 its correct otherwise it doesnt exist find the boundry shade on the correct side of said boundry

No seriously, if you're gonna waste your time, at least waste it trying to suck your own dick. If you can manage to do that, at least you will have picked up a useful skill. Graphing for pre-calc...? That's just time you will never get back and you'll wonder one day why you were too much of a blubbering vagina to stand up and say you didn't want to waste your time.